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Letnbe a positive integer that is not a perfect square. Prove that √nis irrational.

I understand the basic outline that a proof would have. Assume √nis rational and use a proof by contradiction. We can set √n=p/q where p and q are integers with gcd(p,q)=1. Nown=p^{2}/q^{2}. Next,nq^{2}=p^{2}. This implies p^{2}is divisible by q^{2}, which subsequently implies that p is divisible by q. Ifnwas a perfect square, its root would be an integer so q=1 and this is satisfied. However, ifnis not a perfect square, its root would not be an integer. Thus, p divisible by q shows that gcd(p,q)≠1 and we have a contradiction.

My confusion is with the following step: p^{2}is divisible by q^{2}implies that p is divisible by q. Thank you so much for all of the help you can give me.

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# Homework Help: Real Analysis: Proof of irrationality of all Sqrt[n] where n is not a perfect square

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